multivariable calculus question norm

Question

Given vector space C([a,b],$ \mathbb{R} $) of continuous functions of [a,b] in $ \mathbb{R}. $ Prove that the function $ \left \| f \right \|_{1}=\int_{a}^{b}\left | f(t) \right |dt $ is a norm. Also prove that given I([a,b],$ \mathbb{R} $) integrable functions of [a,b] in $ \mathbb{R} $ , $ \left \| . \right \|$ is not a norm.

Answer 1

All the axioms for norms are trivially verified except positive definiteness. So let us assume that for some $f \in C:= C([a,b],\mathbb{R})$ that $\|f\|_1 = 0$. We aim to show that $f=0$ on $[a,b]$. But if not, $\exists x_0 \in (a,b)$ such that $f(x_0)=: a>0$ (without loss of generality we can take a "$>$" here; else consider $-f$). But by continuity $\exists \delta >0$ such that $(x_0 - \delta, x_0 + \delta) \subset [a,b]$ and such that

$$ \forall x \in (x_0 - \delta, x_0 + \delta) \quad f(x) > \frac{a}{2} .$$

But then $\|f\|_1 \geq 2\delta \cdot \frac{a}{2} > 0$, a condtradiction.

To see that $\| \cdot \|$ isn't a norm on $I$, just note that the function $f(x)$ defined to equal $0$ on $(a,b]$ and $1$ at $a$ has zero integral, but is nonzero. Hence, $\| \cdot \|$ isn't positive definite on $I$.

NB $\quad$ In practice, we just 'quotient out' functions that differ on at most a null set with respect to the Lebesgue measure. Once we've done this, $\| \cdot \|$ is a norm on $I$.